Assuming that each person makes a shaking partnership just once and no one shakes there own hand,Consider the situation. The eight people are in a line. The first person goes down the line shaking hands with everyone else so he shakes hands 7 times. He then sits down.
The second person shakes hands with everyone left in line, six handshakes, and sits down, He has no need to shake ands with the sitting man as they already shook.
The remainder follow the same procedure, shaking hands with those standing, not shaking with those sitting.
The last man standing shakes with no one as no one is standing and sitsThe sum of hand shakings initiated by each person is 7+6+5+4+3+2+1+0 = 28
15 (15 * 15 - 15)/2 = 105
The answer is 15 people. Each shook hands with 14 others, and there are half that many handshakes (pairs). The total number of pairs (distinct handshakes) within the group is defined by the formula T = [n!/(n-2)!] /2 Given T = 105 we get n!/(n-2)!=210 which implies n(n-1)=210 on solving we get n=15
9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55
28 people each shook hands once...?
"They shook hands" is correct.
200? D:
15 (15 * 15 - 15)/2 = 105
6
6*5/2 = 15
Each handshake involves two people. If everyone shook only once then there were 36 x 2 ie 72 guests.
There were ten people at the party. This is a triangular sequence starting with two people: 1, 3, 6, 10, 15, 21, 28, 36, 45, etc. There's an equation for this. With n people at the party, the number of handshakes is n(n-1)/2.
there would be 26 handshakes if they were all done at once
50*49/2 = 1225
The answer is 21 handshakes because the first person shakes hands with the other 6 people. The second person shakes hands with 5 people because they already shook hands with the first person. The third person shakes hands with 4 people because they already shook hands with the first and second person. The fourth person shakes hands with 3 people because they already shook hands with the first, second, and third. The fifth person shakes hands with 2 people because they already shook hands with the first, second, third, and fourth person. The sixth person shakes hands with the seventh person because the rest have already shaken hands with them. The seventh person doesn't have anyone else to shake hands with. Therefore the answer is 21 handshakes.
91
The answer is 15 people. Each shook hands with 14 others, and there are half that many handshakes (pairs). The total number of pairs (distinct handshakes) within the group is defined by the formula T = [n!/(n-2)!] /2 Given T = 105 we get n!/(n-2)!=210 which implies n(n-1)=210 on solving we get n=15
The correct answer is 21. The first person shakes hands with the other 6 people. The second person shakes hands with 5 people because they already shook hands with the first person. The third person shakes hands with 4 people because they already shook hands with the first and second person. The fourth person shakes hands with 3 people because they already shook hands with the first, second, and third. The fifth person shakes hands with 2 people because they already shook hands with the first, second, third, and fourth person. The sixth person shakes hands with the seventh person because the rest have already shaken hands with them. The seventh person doesn't have anyone else to shake hands with. So the total would be 21...